$SL_2(\mathbf{Z},8\mathbf{Z})$ differs from $E_2(\mathbf{Z},8\mathbf{Z})$. Has this result appeared in the literature? I know a proof that the congruence subgroup $SL_2(\mathbf{Z},8\mathbf{Z})$ differs from its subgroup $E_2(\mathbf{Z},8\mathbf{Z})$, but can't find this fact in the literature.  Does anyone know a citation for this?
 A: In fact, this holds for any $N$ such that $X=\mathbb{H}^2/PSL_2(\mathbb{Z},N\mathbb{Z})$ has genus $>0$. The point is that the normal subgroup generated by parabolics in $SL_2(\mathbb{Z},N\mathbb{Z})$ will generate the kernel of the map obtained by filling in the cusps of the Riemann surface $X$ to get a closed Riemann surface $\hat{X}$. If $\hat{X}$ has genus $\geq 1$, then the kernel will have infinite index (more generally, if there's torsion, the filling will be an orbifold, which will have non-trivial fundamental group if it has genus $>0$ or more than two orbifold cone points, or two cone points of the same order). In fact, $X$ has genus $=5$. 
Actually, the group I've described will contain $E_2(\mathbb{Z},N\mathbb{Z})$, and is the normalizer in $SL_2(\mathbb{Z})$ of a primitive upper triangular element in $SL_2(\mathbb{Z},N\mathbb{Z})$ . If you take the normalizer in $SL_2(\mathbb{Z},N\mathbb{Z})$, you get an even small subgroup obtained by the kernel of filling in a single cusp of $X$. In this case, one need only show that there are at least 3 cusps, or the genus is $>0$, or there is some combination of elliptic points and cusps which makes the filling have non-trivial fundamental group. 
